If 6.46 L of gaseous ethanol reacts with 16.1 L O2, what is the maximum volume o

Question

If 6.46 L of gaseous ethanol reacts with 16.1 L O2, what is the maximum volume of gaseous carbon dioxde produced? Assume that the temperature of the reactants and products is 425 °C and the pressure remains constant at 1.00 atm.

CH3CH2OH(g) + 3 O2(g) ---> 2 CO2(g) + 3 H2O(g)

a) 6.46 L b)10.7 L c)12.9 L (is not correct according to my chem professor) d)16.1 L e) 22.6 L

This was a question on an exam and I am studying for finals to understand the equation. I selected 12.9L and he said that it was incorrect. Please explain and thank you!

Explanation / Answer

C2H5OH + 3 O2 ---> 2CO2 + 3H2O

we find moles of gas by using PV = nRT

P = 1atm , T = 425 C = 425+273 = 698 K , V = 6.46 L

1 atm x 6.46 L = n x 0.08206 liter atm/molK x 698 K

n = 0.1128

Moles of O2 is found by same formula

1 atm x 16.1 L = n x 0.08206 x 698

n= 0.28

by reaction 1 ethanol reacts with 3O2

hence O2 moles needed for 0.1128 ethanol = 3 x0.1128 = 0.3384 but we had only 0.28 moles O2

hence O2 moles are relatively low , hence O2 is limitng reagent

CO2 moles produce = ( 2/3) O2 moles = ( 2/3) x 0.28 = 0.1867

we find Vol using PV = nRT

1atm x V = 0.1867 x 0.08206 x 698

V = 10.7 L

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